3.4. RELIABILITY INDEX ˇ 123
Solution:
Based on the provided information in this problem, the component strength index is the ultimate
material strength S
u
and the component stress index is the maximum stress
max
at the critical
section. e limit state function of this component will be:
g
.
S
u
;
max
/
D S
u
max
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
D 0 Limit state
< 0 Failure:
Per the provided information, the distribution parameters of the exponentially distributed ulti-
mate strength, and the exponentially distributed maximum stress are:
S
u
D
1
S
u
D
1
62:5
D 0:016
max
D
1
max
D
1
12:4
D
0:081:
Per Equation (3.20), the reliability of this component is:
R D P
Œ
g
.
S
u
;
max
/
0
D
max
S
u
C
max
D
0:081
0:016 C0:081
D 0:835:
3.4 RELIABILITY INDEX ˇ
e reliability index ˇ has been mentioned in Section 3.2 and defined by Equation (3.12):
ˇ D
g
g
D
S
Q
q
2
S
C
2
Q
: (3.12)
e reliability index ˇ will be a frequently used important parameter, which can be used to
calculate the reliability of the component per Equation (3.11):
R D P
Œ
g
.
S; Q
/
0
D ˆ
g
g
D ˆ.ˇ/: (3.11)
Now, we will explain the physical meaning of the reliability index ˇ.
In the limit state function of a component, the component strength index S and the
component stress index Q are random variables and can be functions of other random variables.
For a general case, let use the following general expression of the surface of a limit state function:
g
.
S; Q
/
D S Q D g
.
X
1
; X
2
; : : : ; X
n
/
D 0: (3.21)
124 3. COMPUTATIONAL METHODS FOR THE RELIABILITY OF A COMPONENT
is limit state function can be a linear function and a nonlinear function of all random variables.
If all random variables X
i
.i D 1; 2; : : : ; n/ in the limit state function are statistically indepen-
dent normal distributions, we can convert them into statistically independent standard normally
distributed random variables Z
i
.i D 1; 2; : : : ; n/ by the following equation:
Z
i
D
X
i
X
i
X
i
i D 1; 2; : : : ; n; (3.22)
where
X
i
and
X
i
are the mean and standard deviation of a normally distributed random
variable X
i
. Z
i
is a corresponding standard normal distributed random variable of the normal
distributed random variable X
i
. After these conversions, the surface of the limit state func-
tion (3.21) becomes:
g
.
S; Q
/
D S Q D g
.
X
1
; X
2
; : : : ; X
n
/
D g
1
.
Z
1
; Z
2
; : : : ; Z
n
/
D 0: (3.23)
g
1
.
Z
1
; Z
2
; : : : ; Z
n
/
D 0 is a surface specified in the standard normally distributed space (a
coordinate system), which consists of standard normally distributed random variables Z
i
.i D
1; 2; : : : ; n/.
e reliability index ˇ is the shortest distance [1, 2] from the origin of the standard normally
distributed space to the surface of the limit state function: g
1
.
Z
1
; Z
2
; : : : ; Z
n
/
D 0.
Example 3.8
e limit state function of a component is g
.
S; Q
/
D S Q. Both S and Q are normal dis-
tributions. e component strength index S has a mean
S
and a standard deviation
S
. e
component stress index Q has a mean
Q
and a standard deviation
Q
. Use this limit state func-
tion to verify that the shortest distance between the origin of the standard normally distributed
space and the surface of the limit state function g
.
S; Q
/
D S Q D 0 is the reliability index
ˇ D .
S
Q
/=
q
2
S
C
2
Q
.
Solution:
e surface of the limit state function g
.
S; Q
/
D S Q is:
g
.
S; Q
/
D S Q D 0: (a)
In the S vs. Q coordinate system where the horizontal axis is Q, and the vertical axis is S , the
limit state function will be a straight line, as shown in Figure 3.4. e region on the left side of
the line of the limit state function is safe because of S > Q. e region on the right side of the
line of the limit state function is a failure region because of S < Q.
Let us convert the component strength index S and the component stress index Q into
standard normally distributed random variables Z
S
and Z
Q
:
Z
S
D
S
S
S
; Z
Q
D
Q
Q
Q
: (b)
3.4. RELIABILITY INDEX ˇ 125
Safe Region
Failure Region
S
O
B
β
α
α
A
P
Q
Z
S
Z
Q
g > 0
g < 0
g (Z
S
, Z
Q
) = 0
, 0
μ
S
−
μ
Q
σ
Q
0, −
μ
S
−
μ
Q
σ
S
Figure 3.4: e line of a limit state function with two normal distributed random variables.
Rewrite Equation (b) as:
S D Z
S
S
C
S
; Q D Z
Q
Q
C
Q
: (c)
Plug Equation (c) into Equation (a), we have a different expression of the surface of the same
limit state function:
g
1
Z
S
; Z
Q
D
.
Z
S
S
C
S
/
Z
Q
Q
C
Q
D
0:
(d)
e standard normal distribution space, in this case, is the Z
S
vs. Z
Q
coordinate system. At
point A with Z
QA
D 0, Z
S
can be obtained from Equation (d):
.
Z
SA
S
C
S
/
0
Q
C
Q
D 0: (e)
erefore,
Z
SA
D
S
Q
S
: (f)
So, the coordinate values at the point A in the standard normal distribution space is
0;
S
Q
S
, as shown in Figure 3.4. Repeating the same procedure, we can get the coordinate
values at the point B as
S
Q
Q
; 0
, as shown in Figure 3.4.
e line OP, as shown in Figure 3.4, is normal to the surface (line) of the limit state
function. So, the length of OP will be the shortest distance between the origin O and the line of
the limit state function in the standard normal distribution space. erefore, the length of OP
should be the reliability index ˇ.
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